| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Equations with conjugate of expressions |
| Difficulty | Standard +0.3 This is a straightforward FP1 question testing basic complex number manipulation: rationalizing denominators, expressing in Cartesian form, using argument conditions, and solving equations with conjugates. All techniques are standard with no novel insight required, making it slightly easier than average A-level difficulty. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3w + 7}{5} = \frac{(p - 4i)}{(3 - i)} \times \frac{(3 + i)}{(3 + i)}\) | M1 | Multiplies by \(\frac{(3+i)}{(3+i)}\) or divide by \((9 - 3i)\) then multiply by \(\frac{(9+3i)}{(9+3i)}\) |
| \(= \begin{pmatrix} 3p + 4 \end{pmatrix} / 10 + \begin{pmatrix} p - 12 \end{pmatrix} / 10\) | B1 | Evidence of \((3-i)(3+i) = 10\) or \(3^2 + 1^2\) or \(9 + 3^2\) Rearranges to \(w = ...\) |
| So, \(w = \begin{pmatrix} 3p - 10 \end{pmatrix} / 6 + \begin{pmatrix} p - 12 \end{pmatrix} / 6\) | dM1 | At least one of either the real or imaginary part of \(w\) is correct in any equivalent form. |
| A1 | Correct \(w\) in the form \(a + bi\). Accept \(a + ib\). |
| Answer | Marks | Guidance |
|---|---|---|
| \((3-i)(3w+7) = 5(p-4i)\) | M1 | |
| \(9w + 21 - 3iw - 7i = 5p - 20i\) | ||
| \(w(9 - 3i) = 5p - 21 - 13i\) | M1 | Sets \(w = a + bi\) and equates at least either the real or imaginary part. |
| Real : \(9a + 3b = 5p - 21\) | B1 | \(9a + 3b = 5p - 21\) |
| Imaginary : \(-3a + 9b = -13\) | ||
| \(b = \frac{p - 12}{6}, a = \frac{3p - 10}{6}\) | dM1 | Solves to finds \(a = ...\) and \(b = ...\). At least one of \(a\) or \(b\) is correct in any equivalent form. |
| $w = \begin{pmatrix} 3p - 10 \end{pmatrix} / 6 + \begin{pmatrix} p - 12 \end{pmatrix} / |
**Part (i)(a)**
| $\frac{3w + 7}{5} = \frac{(p - 4i)}{(3 - i)} \times \frac{(3 + i)}{(3 + i)}$ | M1 | Multiplies by $\frac{(3+i)}{(3+i)}$ or divide by $(9 - 3i)$ then multiply by $\frac{(9+3i)}{(9+3i)}$ |
| $= \begin{pmatrix} 3p + 4 \end{pmatrix} / 10 + \begin{pmatrix} p - 12 \end{pmatrix} / 10$ | B1 | Evidence of $(3-i)(3+i) = 10$ or $3^2 + 1^2$ or $9 + 3^2$ Rearranges to $w = ...$ |
| So, $w = \begin{pmatrix} 3p - 10 \end{pmatrix} / 6 + \begin{pmatrix} p - 12 \end{pmatrix} / 6$ | dM1 | At least one of either the real or imaginary part of $w$ is correct in any equivalent form. |
| | A1 | Correct $w$ in the form $a + bi$. Accept $a + ib$. |
**ALT (i)(a)**
| $(3-i)(3w+7) = 5(p-4i)$ | M1 | |
| $9w + 21 - 3iw - 7i = 5p - 20i$ | | |
| $w(9 - 3i) = 5p - 21 - 13i$ | M1 | Sets $w = a + bi$ and equates at least either the real or imaginary part. |
| Real : $9a + 3b = 5p - 21$ | B1 | $9a + 3b = 5p - 21$ |
| Imaginary : $-3a + 9b = -13$ | | |
| $b = \frac{p - 12}{6}, a = \frac{3p - 10}{6}$ | dM1 | Solves to finds $a = ...$ and $b = ...$. At least one of $a$ or $b$ is correct in any equivalent form. |
| $w = \begin{pmatrix} 3p - 10 \end{pmatrix} / 6 + \begin{pmatrix} p - 12 \end{pmatrix} /\begin{enumerate}
\item (i) Given that
\end{enumerate}
$$\frac { 3 w + 7 } { 5 } = \frac { p - 4 \mathrm { i } } { 3 - \mathrm { i } } \quad \text { where } p \text { is a real constant }$$
(a) express $w$ in the form $a + b \mathrm { i }$, where $a$ and $b$ are real constants.
Give your answer in its simplest form in terms of $p$.
Given that arg $w = - \frac { \pi } { 2 }$\\
(b) find the value of $p$.\\
(ii) Given that
$$( z + 1 - 2 i ) ^ { * } = 4 i z$$
find $z$, giving your answer in the form $z = x + i y$, where $x$ and $y$ are real constants.\\
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\hfill \mbox{\textit{Edexcel FP1 2018 Q9 [12]}}